poj 3133 Manhattan Wiring

http://poj.org/problem?id=3133

考虑插头 dp

用四进制表示一个插头的状态，0 表示没有插头，2 表示这个插头是连接两个 2 的，3 同理

然后就是大力分类讨论了

这题还是比较友善的一题，思路相对来说简单很多

我写的括号序列的方法状态是满的，数组必须开到 $ 3 ^ 9 $ 才能过

#include <cstdio>
#include <cstring>
#include <algorithm>
#define CIOS ios::sync_with_stdio(false);
#define rep(i, a, b) for(register int i = a; i <= b; i++)
#define per(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> ";
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename T>
inline void read(T &f) {
	f = 0; T fu = 1; char c = getchar();
	while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); }
	while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
	f *= fu;
}

template <typename T>
void print(T x) {
	if(x < 0) putchar('-'), x = -x;
	if(x < 10) putchar(x + 48);
	else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
	print(x); putchar(t);
}

const int mod = 19687, N = mod + 50, INF = 0x7fffffff;

int a[15][15], bin[15], f[2][N], v[2][N], head[N], tot[2], nxt[N], n, m, now;

// 状态只有 0, 2, 3, 表示没有插头, 2 插头和 3 插头 

void ins(int zt, int val) {
	int u = zt % mod;
	for(register int i = head[u]; i; i = nxt[i])
		if(v[now][i] == zt) {
			f[now][i] = min(f[now][i], val);
			return;
		}
	nxt[++tot[now]] = head[u]; head[u] = tot[now];
	v[now][tot[now]] = zt; f[now][tot[now]] = val;
}

void sol() {
	tot[now] = 1; f[now][1] = v[now][1] = 0;
	for(register int i = 1; i <= n; i++) {
		for(register int i = 1; i <= tot[now]; i++) v[now][i] <<= 2;
		for(register int j = 1; j <= m; j++) {
			now ^= 1; memset(head, 0, sizeof(head)); tot[now] = 0;
			for(register int k = 1; k <= tot[now ^ 1]; k++) {
				int zt = v[now ^ 1][k], val = f[now ^ 1][k];
				int left = (zt >> ((j << 1) - 2)) & 3, up = (zt >> (j << 1)) & 3;
				int right = (j == m) ? 1 : a[i][j + 1], down = (i == n) ? 1 : a[i + 1][j];
				if(a[i][j] == 1) {
					if(!left && !up) ins(zt, val);
				} else if(!left && !up) {
					if(!a[i][j]) {
						ins(zt, val);
						if(right + down == 5 || right == 1 || down == 1) continue;
						if(right == 2 || down == 2) {
							// 当两边有 2 插头时, 向下和向右摆 2 插头
							ins(zt ^ (bin[j - 1] << 1) ^ (bin[j] << 1), val + 1);
						} else if(right == 3 || down == 3) {
							ins(zt ^ (bin[j - 1] * 3) ^ (bin[j] * 3), val + 1);
						} else {
							ins(zt ^ (bin[j - 1] << 1) ^ (bin[j] << 1), val + 1);
							ins(zt ^ (bin[j - 1] * 3) ^ (bin[j] * 3), val + 1);
						}
					} else {
						// 当前位置是一个 2 或 3
						if(a[i][j] + right != 5 && right != 1) {
							ins(zt ^ (bin[j] * a[i][j]), val + 1);
						}
						if(a[i][j] + down != 5 && down != 1) {
							ins(zt ^ (bin[j - 1] * a[i][j]), val + 1);
						}
					}
				} else if(left && up) {
					if(left + up == 5 || a[i][j]) continue;
					ins(zt ^ (bin[j - 1] * left) ^ (bin[j] * up), val + 1);
				} else if(left && !up) {
					if(a[i][j] == 0) {
						if(right == 0 || right == left) ins(zt ^ (bin[j - 1] * left) ^ (bin[j] * left), val + 1);
						if(down == 0 || down == left) ins(zt, val + 1);
					} else if(a[i][j] == left) {
						ins(zt ^ (bin[j - 1] * left), val + 1);
					}
				} else if(!left && up) {
					if(a[i][j] == 0) {
						if(right == 0 || right == up) ins(zt, val + 1);
						if(down == 0 || down == up) ins(zt ^ (bin[j] * up) ^ (bin[j - 1] * up), val + 1);
					} else if(a[i][j] == up) {
						ins(zt ^ (bin[j] * up), val + 1);
					}
				}
			}
		}
	}
}

int main() {
	bin[0] = 1; for(register int i = 1; i <= 11; i++) bin[i] = bin[i - 1] << 2;
	while(scanf("%d %d", &n, &m) == 2 && n && m) {
		memset(a, 0, sizeof(a)); memset(head, 0, sizeof(head)); memset(tot, 0, sizeof(tot));
		for(register int i = 1; i <= n; i++) {
			for(register int j = 1; j <= m; j++) read(a[i][j]);
		}
		sol(); int ans = INF;
		for(register int i = 1; i <= tot[now]; i++) ans = min(ans, f[now][i]);
		if(ans == INF) ans = 2; print(ans - 2, '\n');
	}
	return 0;
}